What is the integral of 4x^2e^{4x} ?

The integral of 4x^2e^{4x} is e^{4x}x^2-1/2 e^{4x}x+1/8 e^{4x}+C

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integral of 4x^2e^{4x}

\int 4 x^{2} e^{4 x} d x

e^{4 x} x^{2} - \frac{1}{2} e^{4 x} x + \frac{1}{8} e^{4 x} + C

Solution steps

\int 4 x^{2} e^{4 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int x^{2} e^{4 x} d x

Apply Integration By Parts

: \frac{1}{4} e^{4 x} x^{2} - \int \frac{1}{2} e^{4 x} x d x

= 4 (\frac{1}{4} e^{4 x} x^{2} - \int \frac{1}{2} e^{4 x} x d x)

\int \frac{1}{2} e^{4 x} x d x = \frac{1}{32} (4 e^{4 x} x - e^{4 x})

= 4 (\frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x}))

Simplify

4 (\frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x})) : e^{4 x} x^{2} - \frac{1}{2} e^{4 x} x + \frac{1}{8} e^{4 x}

= e^{4 x} x^{2} - \frac{1}{2} e^{4 x} x + \frac{1}{8} e^{4 x}

Add a constant to the solution

= e^{4 x} x^{2} - \frac{1}{2} e^{4 x} x + \frac{1}{8} e^{4 x} + C

3 - x

y = x^{8 c o s (x)}

\frac{d}{d x} (\frac{3 x}{1 + x ^{3}})

\frac{\partial}{\partial x} (5 x y e^{- x^{2} y})

What is the integral of 4x^2e^{4x} ?
The integral of 4x^2e^{4x} is e^{4x}x^2-1/2 e^{4x}x+1/8 e^{4x}+C